If the inclination of the resultant to the given line, Ꮎ = tan 0 = P sin a+Q sin ẞ+ R sin y Pcos a+ cos B + R cos y' and the resultant is equal to the square root of P2 + Q2+ R2 + 2 QR cos (B−y) + 2RP cos (y—a) +2PQ cos (a−B). (3) Three forces, each equal to P, act on a point 0, in directions OA, OB, OC; the angle AOC being a right angle, and the line OB bisecting the angle AOC: to find the magnitude of the resultant of the three forces. The resultant force is equal to P(1+√2). (4) Three forces P, QR, act on a point: R acts oppositely to P, and Q at right angles to both: to find the magnitude and direction of the resultant force. The magnitude of the resultant force is equal to {(P− R)2 + Q2}3, and its angle of inclination to P's direction is equal to (5) A particle is placed in the middle point of a fixed horizontal, equilateral, and triangular board, and is kept in equilibrium by three equal weights, which act by means of strings passing through the angular points. Supposing the particle to be moved through a certain space towards one of the angular points, and there fixed; and that the resultant force exerted upon it by the strings is then equal to half of each of the weights, to find the inclinations of the strings to each other. Let ABC be the triangle and suppose the particle to be displaced towards A: then, O being the new position of the particle, the angle AOB or AOC will be equal to CHAPTER IV. EQUILIBRIUM OF ANY NUMBER OF FORCES ACTING ON A POINT. SECT. 1. Geometrical Method. (1) A circle, the plane of which is vertical, has a centre of constant repulsive force, equal to gravity, at one extremity of the horizontal diameter: to find the position of equilibrium of a particle within the circle. Let P, fig. (22), be the position of equilibrium of the particle, A being the centre of force, and C the centre of the circle. Draw PQ vertically to intersect the horizontal diameter in Q. Now, since the particle is acted on by two equal forces, the repulsive force along AP, and the force of gravity along QP, it is necessary to equilibrium that the reaction of the circle, which takes place along PC, bisect the angle between PA and PQ. which determines the position of equilibrium. (2) Shew that within a quadrilateral, no two sides of which are parallel, there is but one point at which forces, acting towards the corners and proportional to the distances of the point from them, can be in equilibrium. Campion and Walton: Solutions of the Cambridge Problems of 1857. W. S. 4 SECT. 2. Components along two lines at right angles to each other. (1) A weight W is sustained on an inclined plane by a certain force; the inclination of the force to the inclined plane is 30°, and the inclination of the plane to the horizon is 30°: to find the pressure on the inclined plane. Let AB, fig. (23), be the inclined plane; let R = the reaction of the plane against the weight, and P= the sustaining force. Then, resolving forces along and perpendicularly to the inclined plane, we have The figure (24) exhibits the two groups into which by resolution we have decomposed the three forces P, W, R; the equations (1) and (2) expressing the conditions for the equilibrium of the two new groups, each regarded as independent of the other. (2) A, B, C, D, fig. (25), are four tacks forming the angles of a square, the lines AB and CD being horizontal: a fine string AEB hangs over the higher tacks, its lowest point E being at the centre of the square: another fine string PDECP, with equal weights P, P, at its ends, hangs over the former string and over the lower tacks: to find the tension of the string AEB and the pressures on all the tacks. Let Tbe the tension of the string AEB. Then the point E is kept at rest by a force T along EA, a force T along EB, a force P along EC, and a force P along ED, that is, by one system of forces in the line AC, and another in the perpendicular line BD: hence T=P. Again, the pressure on either of the higher tacks is equal to (3) A string of given length is attached to the highest point O, fig. (26), of a given sphere: a weight P is fixed to the lower end of the string: to find the pressure on the hemisphere and the tension of the string. Let OB be the arc touched by the string, and C the centre of the sphere: let < BCO = a. L Let R be the reaction of the sphere against the weight; this reaction will take place in the direction CB. Let T be the tension of the string: T will act on the weight tangentially to the sphere. For the equilibrium of the weight we have, resolving forces tangentially and normally, and = T= Psin a, RP cos a. r = Let c the length of the string, and the radius of the sphere: then (4) The ends of a fine cord of given length are tied to two given points in the same horizontal line, and a smooth ring, sliding on the cord, sustains a given weight: to find the tension of the cord. Let A, B, fig. (27), be the two given points, ACB the cord in a position of rest: draw CH vertically to meet the horizontal line AB in H. Let W the given weight, and T = the tension of the string. Let ACH=0, ▲ BCH=4. Then, since the = L ring C is kept at rest by the tensions in the directions CA, CB, and the weight W, we have, resolving horizontally, Let 2c the length of the cord, and let AB=2a: then BC=c, BH= a, and therefore (5) A man, fig. (28), with a perfectly smooth spherical head, wears a conical hat: to find the whole pressure on his head. = Let W the weight and 2a = the vertical angle of his hat, and let R = the reaction of his head against his hat at any point of the circle of contact: this reaction will take place along the radius of his head at this point. Then R sin a will denote the vertical component of R, and the sum of all such components must be equal to W: hence Hence the whole pressure on his head, which is equal to Σ(R), is equal to W sin a COR. If a be exceedingly small, and W finite, the pressure |